`if A[{:(1,3,2),(2,0,3),(1,-1,1):}],`then find `A^(3)-2A^(2)+A-I_(3).`

Explanation

To solve the problem, we need to compute \( A^3 - 2A^2 + A - I_3 \) where \( A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1 \end{pmatrix} \) and \( I_3 \) is the identity matrix of order 3.

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 1*1 + 3*2 + 2*1 = 1 + 6 + 2 = 9 \)
- First row, second column: \( 1*3 + 3*0 + 2*(-1) = 3 + 0 - 2 = 1 \)
- First row, third column: \( 1*2 + 3*3 + 2*1 = 2 + 9 + 2 = 13 \)

- Second row, first column: \( 2*1 + 0*2 + 3*1 = 2 + 0 + 3 = 5 \)
- Second row, second column: \( 2*3 + 0*0 + 3*(-1) = 6 + 0 - 3 = 3 \)
- Second row, third column: \( 2*2 + 0*3 + 3*1 = 4 + 0 + 3 = 7 \)

- Third row, first column: \( 1*1 + (-1)*2 + 1*1 = 1 - 2 + 1 = 0 \)
- Third row, second column: \( 1*3 + (-1)*0 + 1*(-1) = 3 + 0 - 1 = 2 \)
- Third row, third column: \( 1*2 + (-1)*3 + 1*1 = 2 - 3 + 1 = 0 \)

Thus,

\[ A^2 = \begin{pmatrix} 9 & 1 & 13 \\ 5 & 3 & 7 \\ 0 & 2 & 0 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

Now, we compute \( A^3 = A^2 \cdot A \):

\[ A^3 = \begin{pmatrix} 9 & 1 & 13 \\ 5 & 3 & 7 \\ 0 & 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 9*1 + 1*2 + 13*1 = 9 + 2 + 13 = 24 \)
- First row, second column: \( 9*3 + 1*0 + 13*(-1) = 27 + 0 - 13 = 14 \)
- First row, third column: \( 9*2 + 1*3 + 13*1 = 18 + 3 + 13 = 34 \)

- Second row, first column: \( 5*1 + 3*2 + 7*1 = 5 + 6 + 7 = 18 \)
- Second row, second column: \( 5*3 + 3*0 + 7*(-1) = 15 + 0 - 7 = 8 \)
- Second row, third column: \( 5*2 + 3*3 + 7*1 = 10 + 9 + 7 = 26 \)

- Third row, first column: \( 0*1 + 2*2 + 0*1 = 0 + 4 + 0 = 4 \)
- Third row, second column: \( 0*3 + 2*0 + 0*(-1) = 0 + 0 + 0 = 0 \)
- Third row, third column: \( 0*2 + 2*3 + 0*1 = 0 + 6 + 0 = 6 \)

Thus,

\[ A^3 = \begin{pmatrix} 24 & 14 & 34 \\ 18 & 8 & 26 \\ 4 & 0 & 6 \end{pmatrix} \]

### Step 3: Calculate \( 2A^2 \)

Now, we calculate \( 2A^2 \):

\[ 2A^2 = 2 \cdot \begin{pmatrix} 9 & 1 & 13 \\ 5 & 3 & 7 \\ 0 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 18 & 2 & 26 \\ 10 & 6 & 14 \\ 0 & 4 & 0 \end{pmatrix} \]

### Step 4: Calculate \( A - I_3 \)

Next, we find \( A - I_3 \):

\[ I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Thus,

\[ A - I_3 = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 3 & 2 \\ 2 & -1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]

### Step 5: Combine all results

Now we can combine all the results to find \( A^3 - 2A^2 + A - I_3 \):

\[ A^3 - 2A^2 + A - I_3 = \begin{pmatrix} 24 & 14 & 34 \\ 18 & 8 & 26 \\ 4 & 0 & 6 \end{pmatrix} - \begin{pmatrix} 18 & 2 & 26 \\ 10 & 6 & 14 \\ 0 & 4 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 3 & 2 \\ 2 & -1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]

Calculating each element:

- First row: \( (24 - 18 + 0), (14 - 2 + 3), (34 - 26 + 2) = (6, 15, 10) \)
- Second row: \( (18 - 10 + 2), (8 - 6 - 1), (26 - 14 + 3) = (10, 1, 15) \)
- Third row: \( (4 - 0 + 1), (0 - 4 - 1), (6 - 0 + 0) = (5, -5, 6) \)

Thus, the final result is:

\[ A^3 - 2A^2 + A - I_3 = \begin{pmatrix} 6 & 15 & 10 \\ 10 & 1 & 15 \\ 5 & -5 & 6 \end{pmatrix} \]